94 29. COMPACT 2-DIMENSIONAL ANCIENT SOLUTIONS
assume that s(<p(f3z(t)));:::: s for all t E [1, £ 1 ). Define L2 to be the first number t
for which s(<p(f3z(t))) = c^1 - 4.
BP(lOO)
' y ' '
"
'
/z ' ,' ' • p
FIGURE 29.3.
Note that o:~(uz) is perpendicular to /3~(0) with respect to g. By choosing
c E (0, c5/40) small enough, from the geometry of a normalized c-neck and since
/3zl(o,L 2 ] is a minimal geodesic with L2 large, we have that
1/3~(0) - (<p-l)* ( :s) lg ~ :o.
By considering vectors perpendicular to those on the LHS, we obtain that
(29.78) lo:~(uz) - (<p-^1 )* (:e) lg~ :o.
By (29.78), the curve 0: 8 can be parametrized in the neck coordinates by
(s (B), B), e E 51. Moreover, at any smooth point of a 8 , the tangent vector
a~(e) = (s' (B), 1) satisfies Is' (B)I ~ 185. Hence Lgcy 1 (o:s) = f 51 lo:~(B)I de is ~-close
to 2n.
On the other hand, since we are in an c-neck, L 9 (rsy) is also ~-close to 2n. D
Now we consider an ancient solution (5^2 ,g(t)). We first prove that areas of
balls are bounded from below linearly in their radii.
LEMMA 29.27 (Area lower bound for balls). There exists a constant c 0 > 0
such that
(29.79) Areag(t)(B$(t)(r));:::: c 0 r
for any (p,t) E 52 x (-oo,-1] and 3 ~ r ~ diam(g(t)).
PROOF. Recall that by Rg(t) ;:::: 0 and the relative Bishop- Gromov volume
comparison theorem, we have for x E 52 that
(29.80)
Areag(t)(Bi(t)(r4) - Bi(t)(r3)) < r~ - r§
Areag(t)(Bi(t)(r2) - Bi(t\r 1 )) - r~ -ri